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PHXI14:OSCILLATIONS

364111 A particle is vibrating in a simple harmonic motion with an amplitude of \(4\;cm\). At what displacement from the equilibrium position, is its energy half potential and half kinetic

1 \(1\;cm\)

2 \(\sqrt 2 \;cm\)

3 \(3\;cm\)

4 \(2\sqrt 2 \;cm\)

PHXI14:OSCILLATIONS

364112 The quantity which does not vary periodically for a particle performing SHM is

1 Acceleration

2 Total energy

3 Displacement

4 Velocity

PHXI14:OSCILLATIONS

364113 A long spring, when stretched by a distance \(x\), has potential energy \(U\). On increasing the stretching to \(n x\), the potential energy of the spring will be

1 \(\dfrac{U}{n}\)

2 \(n U\)

3 \(n^{2} U\)

4 \(\dfrac{U}{n^{2}}\)

PHXI14:OSCILLATIONS

364114 A particle doing \(S H M\) having amplitude \(5\;cm,\) mass \(0.5\;kg\) and angular frequency \(5\,rad/s\) is at \(1\;cm\) from mean position. Find potential energy and kinetic energy

1 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 150 \times {10^{ - 3}}\;J\)

2 \(KE = 150 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

3 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

4 \(KE = 150 \times {10^{ - 3}}\;J,PE = 150 \times {10^{ - 4}}\;J\)

PHXI14:OSCILLATIONS

364115 A particle of mass \(m\) is located in a undimensional potential field where potential energy of the particle depends on the coordinate \(x\) as \(U(x)=\dfrac{A}{x^{2}}-\dfrac{B}{x}\) where \(A\) and \(B\) are positive contacts. Find the time period of small oscillations that the particle will perform about equilibrium position.

1 \(4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

2 \(2 \pi A \sqrt{\dfrac{m A}{B A}}\)

3 \(8 \pi A \sqrt{\dfrac{m A}{2 B A}}\)

4 \(\pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

Explanation:

The potential energy function for the mass \(m\) in a undimensional potential field is given as
\(U(x) = \frac{A}{{{x^2}}} - \frac{B}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Since this is the potential energy of mass \(m\) in a potential field, so the force acting on the mass is obtained by the relation:
\(F = - \frac{{dU}}{{dx}} = \frac{{2A}}{{{x^3}}} - \frac{B}{{{x^2}}}\,\,\,\,\,\,\,\,\,\,(2)\)
Now, the equilibrium position is given by
\(\begin{aligned}& F=0 \text { or } \dfrac{2 A}{x^{3}}-\dfrac{B}{x^{2}}=0 \\& (2 A-B x)=0\end{aligned}\)
So, \(\quad x=\dfrac{2 A}{B}\)
Since \(x=2 A / B\) gives stable situation
(because energy is minimum here; \(d^{2} U / d x^{2}\) is positive), so stable equilibrium position is given as \(x=\dfrac{2 A}{B}\)
Now if the mass is displaced slightly away from the equilibrium position, the force acting on mass \(m\) is given as
\(F=\dfrac{2 A}{[2 A / B+\Delta x]^{3}}-\dfrac{B}{[2 A / B+\Delta x]^{2}}\)
\(\begin{aligned}& =\dfrac{2 A}{\left(\dfrac{2 A}{B}\right)^{3}\left(1+\dfrac{B \Delta x}{2 A}\right)^{3}}-\dfrac{B}{\left(\dfrac{2 A}{B}\right)^{2}\left(1+\dfrac{B \Delta x}{2 A}\right)^{2}} \\& =\dfrac{2 A}{8 A^{3} / B^{3}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-3}-\dfrac{B}{4 A^{2} / B^{2}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-2}\end{aligned}\)
So, using binomial expansion for small \(\Delta x\), we get
\(\begin{aligned}F & =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}\right]-\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{2 B \Delta x}{2 A}\right] \\& =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}-1+\dfrac{2 B \Delta x}{2 A}\right]=-\left(\dfrac{B^{4}}{8 A^{3}}\right) \Delta x\end{aligned}\)
We can write
\(m\frac{{{d^2}x}}{{d{t^2}}} = - \frac{{{B^4}}}{{8{A^3}}}\Delta x\)
\(\frac{{{d^2}x}}{{d{t^2}}} = - \left( {\frac{{{B^4}}}{{8m{A^3}}}} \right)\Delta x\,\,\,\,\,\,\,\,\,\,(3)\)
\(\frac{{{d^2}x}}{{d{t^2}}} \propto - \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)\)
So from the above relations, it is clear that particle will perform SHM. But for any SHM,
\(\dfrac{d^{2} x}{d t^{2}}=-\omega^{2} \Delta x\)
So, from eqs. (3) and (4), we have \(\omega^{2}=\dfrac{B^{4}}{8 m A^{3}}\)
\(T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{8 A^{3} m}{B^{4}}}=4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

PHXI14:OSCILLATIONS

364111 A particle is vibrating in a simple harmonic motion with an amplitude of \(4\;cm\). At what displacement from the equilibrium position, is its energy half potential and half kinetic

1 \(1\;cm\)

2 \(\sqrt 2 \;cm\)

3 \(3\;cm\)

4 \(2\sqrt 2 \;cm\)

PHXI14:OSCILLATIONS

364112 The quantity which does not vary periodically for a particle performing SHM is

1 Acceleration

2 Total energy

3 Displacement

4 Velocity

PHXI14:OSCILLATIONS

364113 A long spring, when stretched by a distance \(x\), has potential energy \(U\). On increasing the stretching to \(n x\), the potential energy of the spring will be

1 \(\dfrac{U}{n}\)

2 \(n U\)

3 \(n^{2} U\)

4 \(\dfrac{U}{n^{2}}\)

PHXI14:OSCILLATIONS

364114 A particle doing \(S H M\) having amplitude \(5\;cm,\) mass \(0.5\;kg\) and angular frequency \(5\,rad/s\) is at \(1\;cm\) from mean position. Find potential energy and kinetic energy

1 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 150 \times {10^{ - 3}}\;J\)

2 \(KE = 150 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

3 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

4 \(KE = 150 \times {10^{ - 3}}\;J,PE = 150 \times {10^{ - 4}}\;J\)

PHXI14:OSCILLATIONS

364115 A particle of mass \(m\) is located in a undimensional potential field where potential energy of the particle depends on the coordinate \(x\) as \(U(x)=\dfrac{A}{x^{2}}-\dfrac{B}{x}\) where \(A\) and \(B\) are positive contacts. Find the time period of small oscillations that the particle will perform about equilibrium position.

1 \(4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

2 \(2 \pi A \sqrt{\dfrac{m A}{B A}}\)

3 \(8 \pi A \sqrt{\dfrac{m A}{2 B A}}\)

4 \(\pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

Explanation:

The potential energy function for the mass \(m\) in a undimensional potential field is given as
\(U(x) = \frac{A}{{{x^2}}} - \frac{B}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Since this is the potential energy of mass \(m\) in a potential field, so the force acting on the mass is obtained by the relation:
\(F = - \frac{{dU}}{{dx}} = \frac{{2A}}{{{x^3}}} - \frac{B}{{{x^2}}}\,\,\,\,\,\,\,\,\,\,(2)\)
Now, the equilibrium position is given by
\(\begin{aligned}& F=0 \text { or } \dfrac{2 A}{x^{3}}-\dfrac{B}{x^{2}}=0 \\& (2 A-B x)=0\end{aligned}\)
So, \(\quad x=\dfrac{2 A}{B}\)
Since \(x=2 A / B\) gives stable situation
(because energy is minimum here; \(d^{2} U / d x^{2}\) is positive), so stable equilibrium position is given as \(x=\dfrac{2 A}{B}\)
Now if the mass is displaced slightly away from the equilibrium position, the force acting on mass \(m\) is given as
\(F=\dfrac{2 A}{[2 A / B+\Delta x]^{3}}-\dfrac{B}{[2 A / B+\Delta x]^{2}}\)
\(\begin{aligned}& =\dfrac{2 A}{\left(\dfrac{2 A}{B}\right)^{3}\left(1+\dfrac{B \Delta x}{2 A}\right)^{3}}-\dfrac{B}{\left(\dfrac{2 A}{B}\right)^{2}\left(1+\dfrac{B \Delta x}{2 A}\right)^{2}} \\& =\dfrac{2 A}{8 A^{3} / B^{3}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-3}-\dfrac{B}{4 A^{2} / B^{2}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-2}\end{aligned}\)
So, using binomial expansion for small \(\Delta x\), we get
\(\begin{aligned}F & =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}\right]-\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{2 B \Delta x}{2 A}\right] \\& =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}-1+\dfrac{2 B \Delta x}{2 A}\right]=-\left(\dfrac{B^{4}}{8 A^{3}}\right) \Delta x\end{aligned}\)
We can write
\(m\frac{{{d^2}x}}{{d{t^2}}} = - \frac{{{B^4}}}{{8{A^3}}}\Delta x\)
\(\frac{{{d^2}x}}{{d{t^2}}} = - \left( {\frac{{{B^4}}}{{8m{A^3}}}} \right)\Delta x\,\,\,\,\,\,\,\,\,\,(3)\)
\(\frac{{{d^2}x}}{{d{t^2}}} \propto - \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)\)
So from the above relations, it is clear that particle will perform SHM. But for any SHM,
\(\dfrac{d^{2} x}{d t^{2}}=-\omega^{2} \Delta x\)
So, from eqs. (3) and (4), we have \(\omega^{2}=\dfrac{B^{4}}{8 m A^{3}}\)
\(T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{8 A^{3} m}{B^{4}}}=4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

PHXI14:OSCILLATIONS

364111 A particle is vibrating in a simple harmonic motion with an amplitude of \(4\;cm\). At what displacement from the equilibrium position, is its energy half potential and half kinetic

1 \(1\;cm\)

2 \(\sqrt 2 \;cm\)

3 \(3\;cm\)

4 \(2\sqrt 2 \;cm\)

PHXI14:OSCILLATIONS

364112 The quantity which does not vary periodically for a particle performing SHM is

1 Acceleration

2 Total energy

3 Displacement

4 Velocity

PHXI14:OSCILLATIONS

364113 A long spring, when stretched by a distance \(x\), has potential energy \(U\). On increasing the stretching to \(n x\), the potential energy of the spring will be

1 \(\dfrac{U}{n}\)

2 \(n U\)

3 \(n^{2} U\)

4 \(\dfrac{U}{n^{2}}\)

PHXI14:OSCILLATIONS

364114 A particle doing \(S H M\) having amplitude \(5\;cm,\) mass \(0.5\;kg\) and angular frequency \(5\,rad/s\) is at \(1\;cm\) from mean position. Find potential energy and kinetic energy

1 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 150 \times {10^{ - 3}}\;J\)

2 \(KE = 150 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

3 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

4 \(KE = 150 \times {10^{ - 3}}\;J,PE = 150 \times {10^{ - 4}}\;J\)

PHXI14:OSCILLATIONS

364115 A particle of mass \(m\) is located in a undimensional potential field where potential energy of the particle depends on the coordinate \(x\) as \(U(x)=\dfrac{A}{x^{2}}-\dfrac{B}{x}\) where \(A\) and \(B\) are positive contacts. Find the time period of small oscillations that the particle will perform about equilibrium position.

1 \(4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

2 \(2 \pi A \sqrt{\dfrac{m A}{B A}}\)

3 \(8 \pi A \sqrt{\dfrac{m A}{2 B A}}\)

4 \(\pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

Explanation:

The potential energy function for the mass \(m\) in a undimensional potential field is given as
\(U(x) = \frac{A}{{{x^2}}} - \frac{B}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Since this is the potential energy of mass \(m\) in a potential field, so the force acting on the mass is obtained by the relation:
\(F = - \frac{{dU}}{{dx}} = \frac{{2A}}{{{x^3}}} - \frac{B}{{{x^2}}}\,\,\,\,\,\,\,\,\,\,(2)\)
Now, the equilibrium position is given by
\(\begin{aligned}& F=0 \text { or } \dfrac{2 A}{x^{3}}-\dfrac{B}{x^{2}}=0 \\& (2 A-B x)=0\end{aligned}\)
So, \(\quad x=\dfrac{2 A}{B}\)
Since \(x=2 A / B\) gives stable situation
(because energy is minimum here; \(d^{2} U / d x^{2}\) is positive), so stable equilibrium position is given as \(x=\dfrac{2 A}{B}\)
Now if the mass is displaced slightly away from the equilibrium position, the force acting on mass \(m\) is given as
\(F=\dfrac{2 A}{[2 A / B+\Delta x]^{3}}-\dfrac{B}{[2 A / B+\Delta x]^{2}}\)
\(\begin{aligned}& =\dfrac{2 A}{\left(\dfrac{2 A}{B}\right)^{3}\left(1+\dfrac{B \Delta x}{2 A}\right)^{3}}-\dfrac{B}{\left(\dfrac{2 A}{B}\right)^{2}\left(1+\dfrac{B \Delta x}{2 A}\right)^{2}} \\& =\dfrac{2 A}{8 A^{3} / B^{3}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-3}-\dfrac{B}{4 A^{2} / B^{2}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-2}\end{aligned}\)
So, using binomial expansion for small \(\Delta x\), we get
\(\begin{aligned}F & =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}\right]-\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{2 B \Delta x}{2 A}\right] \\& =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}-1+\dfrac{2 B \Delta x}{2 A}\right]=-\left(\dfrac{B^{4}}{8 A^{3}}\right) \Delta x\end{aligned}\)
We can write
\(m\frac{{{d^2}x}}{{d{t^2}}} = - \frac{{{B^4}}}{{8{A^3}}}\Delta x\)
\(\frac{{{d^2}x}}{{d{t^2}}} = - \left( {\frac{{{B^4}}}{{8m{A^3}}}} \right)\Delta x\,\,\,\,\,\,\,\,\,\,(3)\)
\(\frac{{{d^2}x}}{{d{t^2}}} \propto - \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)\)
So from the above relations, it is clear that particle will perform SHM. But for any SHM,
\(\dfrac{d^{2} x}{d t^{2}}=-\omega^{2} \Delta x\)
So, from eqs. (3) and (4), we have \(\omega^{2}=\dfrac{B^{4}}{8 m A^{3}}\)
\(T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{8 A^{3} m}{B^{4}}}=4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

PHXI14:OSCILLATIONS

364111 A particle is vibrating in a simple harmonic motion with an amplitude of \(4\;cm\). At what displacement from the equilibrium position, is its energy half potential and half kinetic

1 \(1\;cm\)

2 \(\sqrt 2 \;cm\)

3 \(3\;cm\)

4 \(2\sqrt 2 \;cm\)

PHXI14:OSCILLATIONS

364112 The quantity which does not vary periodically for a particle performing SHM is

1 Acceleration

2 Total energy

3 Displacement

4 Velocity

PHXI14:OSCILLATIONS

364113 A long spring, when stretched by a distance \(x\), has potential energy \(U\). On increasing the stretching to \(n x\), the potential energy of the spring will be

1 \(\dfrac{U}{n}\)

2 \(n U\)

3 \(n^{2} U\)

4 \(\dfrac{U}{n^{2}}\)

PHXI14:OSCILLATIONS

364114 A particle doing \(S H M\) having amplitude \(5\;cm,\) mass \(0.5\;kg\) and angular frequency \(5\,rad/s\) is at \(1\;cm\) from mean position. Find potential energy and kinetic energy

1 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 150 \times {10^{ - 3}}\;J\)

2 \(KE = 150 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

3 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

4 \(KE = 150 \times {10^{ - 3}}\;J,PE = 150 \times {10^{ - 4}}\;J\)

PHXI14:OSCILLATIONS

364115 A particle of mass \(m\) is located in a undimensional potential field where potential energy of the particle depends on the coordinate \(x\) as \(U(x)=\dfrac{A}{x^{2}}-\dfrac{B}{x}\) where \(A\) and \(B\) are positive contacts. Find the time period of small oscillations that the particle will perform about equilibrium position.

1 \(4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

2 \(2 \pi A \sqrt{\dfrac{m A}{B A}}\)

3 \(8 \pi A \sqrt{\dfrac{m A}{2 B A}}\)

4 \(\pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

Explanation:

The potential energy function for the mass \(m\) in a undimensional potential field is given as
\(U(x) = \frac{A}{{{x^2}}} - \frac{B}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Since this is the potential energy of mass \(m\) in a potential field, so the force acting on the mass is obtained by the relation:
\(F = - \frac{{dU}}{{dx}} = \frac{{2A}}{{{x^3}}} - \frac{B}{{{x^2}}}\,\,\,\,\,\,\,\,\,\,(2)\)
Now, the equilibrium position is given by
\(\begin{aligned}& F=0 \text { or } \dfrac{2 A}{x^{3}}-\dfrac{B}{x^{2}}=0 \\& (2 A-B x)=0\end{aligned}\)
So, \(\quad x=\dfrac{2 A}{B}\)
Since \(x=2 A / B\) gives stable situation
(because energy is minimum here; \(d^{2} U / d x^{2}\) is positive), so stable equilibrium position is given as \(x=\dfrac{2 A}{B}\)
Now if the mass is displaced slightly away from the equilibrium position, the force acting on mass \(m\) is given as
\(F=\dfrac{2 A}{[2 A / B+\Delta x]^{3}}-\dfrac{B}{[2 A / B+\Delta x]^{2}}\)
\(\begin{aligned}& =\dfrac{2 A}{\left(\dfrac{2 A}{B}\right)^{3}\left(1+\dfrac{B \Delta x}{2 A}\right)^{3}}-\dfrac{B}{\left(\dfrac{2 A}{B}\right)^{2}\left(1+\dfrac{B \Delta x}{2 A}\right)^{2}} \\& =\dfrac{2 A}{8 A^{3} / B^{3}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-3}-\dfrac{B}{4 A^{2} / B^{2}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-2}\end{aligned}\)
So, using binomial expansion for small \(\Delta x\), we get
\(\begin{aligned}F & =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}\right]-\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{2 B \Delta x}{2 A}\right] \\& =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}-1+\dfrac{2 B \Delta x}{2 A}\right]=-\left(\dfrac{B^{4}}{8 A^{3}}\right) \Delta x\end{aligned}\)
We can write
\(m\frac{{{d^2}x}}{{d{t^2}}} = - \frac{{{B^4}}}{{8{A^3}}}\Delta x\)
\(\frac{{{d^2}x}}{{d{t^2}}} = - \left( {\frac{{{B^4}}}{{8m{A^3}}}} \right)\Delta x\,\,\,\,\,\,\,\,\,\,(3)\)
\(\frac{{{d^2}x}}{{d{t^2}}} \propto - \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)\)
So from the above relations, it is clear that particle will perform SHM. But for any SHM,
\(\dfrac{d^{2} x}{d t^{2}}=-\omega^{2} \Delta x\)
So, from eqs. (3) and (4), we have \(\omega^{2}=\dfrac{B^{4}}{8 m A^{3}}\)
\(T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{8 A^{3} m}{B^{4}}}=4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

PHXI14:OSCILLATIONS

364111 A particle is vibrating in a simple harmonic motion with an amplitude of \(4\;cm\). At what displacement from the equilibrium position, is its energy half potential and half kinetic

1 \(1\;cm\)

2 \(\sqrt 2 \;cm\)

3 \(3\;cm\)

4 \(2\sqrt 2 \;cm\)

PHXI14:OSCILLATIONS

364112 The quantity which does not vary periodically for a particle performing SHM is

1 Acceleration

2 Total energy

3 Displacement

4 Velocity

PHXI14:OSCILLATIONS

364113 A long spring, when stretched by a distance \(x\), has potential energy \(U\). On increasing the stretching to \(n x\), the potential energy of the spring will be

1 \(\dfrac{U}{n}\)

2 \(n U\)

3 \(n^{2} U\)

4 \(\dfrac{U}{n^{2}}\)

PHXI14:OSCILLATIONS

364114 A particle doing \(S H M\) having amplitude \(5\;cm,\) mass \(0.5\;kg\) and angular frequency \(5\,rad/s\) is at \(1\;cm\) from mean position. Find potential energy and kinetic energy

1 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 150 \times {10^{ - 3}}\;J\)

2 \(KE = 150 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

3 \(KE = 6.25 \times {10^{ - 4}}\;J,PE = 6.25 \times {10^{ - 4}}\;J\)

4 \(KE = 150 \times {10^{ - 3}}\;J,PE = 150 \times {10^{ - 4}}\;J\)

PHXI14:OSCILLATIONS

364115 A particle of mass \(m\) is located in a undimensional potential field where potential energy of the particle depends on the coordinate \(x\) as \(U(x)=\dfrac{A}{x^{2}}-\dfrac{B}{x}\) where \(A\) and \(B\) are positive contacts. Find the time period of small oscillations that the particle will perform about equilibrium position.

1 \(4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

2 \(2 \pi A \sqrt{\dfrac{m A}{B A}}\)

3 \(8 \pi A \sqrt{\dfrac{m A}{2 B A}}\)

4 \(\pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

Explanation:

The potential energy function for the mass \(m\) in a undimensional potential field is given as
\(U(x) = \frac{A}{{{x^2}}} - \frac{B}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Since this is the potential energy of mass \(m\) in a potential field, so the force acting on the mass is obtained by the relation:
\(F = - \frac{{dU}}{{dx}} = \frac{{2A}}{{{x^3}}} - \frac{B}{{{x^2}}}\,\,\,\,\,\,\,\,\,\,(2)\)
Now, the equilibrium position is given by
\(\begin{aligned}& F=0 \text { or } \dfrac{2 A}{x^{3}}-\dfrac{B}{x^{2}}=0 \\& (2 A-B x)=0\end{aligned}\)
So, \(\quad x=\dfrac{2 A}{B}\)
Since \(x=2 A / B\) gives stable situation
(because energy is minimum here; \(d^{2} U / d x^{2}\) is positive), so stable equilibrium position is given as \(x=\dfrac{2 A}{B}\)
Now if the mass is displaced slightly away from the equilibrium position, the force acting on mass \(m\) is given as
\(F=\dfrac{2 A}{[2 A / B+\Delta x]^{3}}-\dfrac{B}{[2 A / B+\Delta x]^{2}}\)
\(\begin{aligned}& =\dfrac{2 A}{\left(\dfrac{2 A}{B}\right)^{3}\left(1+\dfrac{B \Delta x}{2 A}\right)^{3}}-\dfrac{B}{\left(\dfrac{2 A}{B}\right)^{2}\left(1+\dfrac{B \Delta x}{2 A}\right)^{2}} \\& =\dfrac{2 A}{8 A^{3} / B^{3}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-3}-\dfrac{B}{4 A^{2} / B^{2}}\left[1+\dfrac{B \Delta x}{2 A}\right]^{-2}\end{aligned}\)
So, using binomial expansion for small \(\Delta x\), we get
\(\begin{aligned}F & =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}\right]-\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{2 B \Delta x}{2 A}\right] \\& =\dfrac{B^{3}}{4 A^{2}}\left[1-\dfrac{3 B \Delta x}{2 A}-1+\dfrac{2 B \Delta x}{2 A}\right]=-\left(\dfrac{B^{4}}{8 A^{3}}\right) \Delta x\end{aligned}\)
We can write
\(m\frac{{{d^2}x}}{{d{t^2}}} = - \frac{{{B^4}}}{{8{A^3}}}\Delta x\)
\(\frac{{{d^2}x}}{{d{t^2}}} = - \left( {\frac{{{B^4}}}{{8m{A^3}}}} \right)\Delta x\,\,\,\,\,\,\,\,\,\,(3)\)
\(\frac{{{d^2}x}}{{d{t^2}}} \propto - \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)\)
So from the above relations, it is clear that particle will perform SHM. But for any SHM,
\(\dfrac{d^{2} x}{d t^{2}}=-\omega^{2} \Delta x\)
So, from eqs. (3) and (4), we have \(\omega^{2}=\dfrac{B^{4}}{8 m A^{3}}\)
\(T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{8 A^{3} m}{B^{4}}}=4 \pi A \sqrt{\dfrac{2 m A}{B^{4}}}\)

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